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CIE 9709 2022 June Paper 62 Q6

A Level / CIE / S2

CIE 9709 2022 June Paper 62 Paper · Question 6

题目

Problem

The masses, in kilograms, of large and small sacks of grain have the distributions N(53,11)\mathrm{N}(53,11) and N(14,3)\mathrm{N}(14,3) respectively.

(a) Find the probability that the mass of a randomly chosen large sack is greater than four times the mass of a randomly chosen small sack.

(5)

(b) A lift can safely carry a maximum mass of 1000 kg1000\,\mathrm{kg}.

Find the probability that the lift can safely carry 1212 randomly chosen large sacks and 2525 randomly chosen small sacks.

(5)
题目中文翻译

大袋谷物和小袋谷物的质量（以千克为单位）分别服从 N(53,11)\mathrm{N}(53,11) 和 N(14,3)\mathrm{N}(14,3) 分布。

(a) 随机选取一大袋和一小袋谷物，求大袋的质量大于小袋质量四倍的概率。

(5)

(b) 一部电梯的最大安全载重为 1000 kg1000\,\mathrm{kg}。

随机选取 1212 大袋和 2525 小袋谷物，求电梯能够安全运载这些谷物的概率。

(5)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。

解答

(a)

解法一

思路

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大袋减4倍同一个小袋，建立正态差分布，方差中小袋项乘16。

答题过程

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Let LL and SS be the masses of the randomly chosen large and small sacks. Then

D=L−4S.D=L-4S.

Then

D∼ N(53−4(14), 11+42(3))= N(−3,59).\begin{align*} D\sim&\,\mathrm{N}(53-4(14),\ 11+4^2(3))\\[4mm] =&\,\mathrm{N}(-3,59). \end{align*}

Therefore

P(L>4S)= 1−Φ(359)= 0.348(3 s.f.).\begin{align*} \mathrm{P}(L>4S)=&\,1-\Phi\bigg(\frac{3}{\sqrt{59}}\bigg)\\[4mm] =&\,\boxed{0.348}\quad\text{(3 s.f.).} \end{align*}

(b)

解法一

思路

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37个随机袋子的总质量均值与方差分别相加。安全即不超过1000kg，连续分布无须连续性修正。

答题过程

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Let TT be the total mass of the 12 large and 25 small randomly selected sacks. Adding the independent sampled masses,

E(T)= 12(53)+25(14)=986,Var(T)= 12(11)+25(3)=207.\begin{align*} \mathrm{E}(T)=&\,12(53)+25(14)=986,\\[4mm] \mathrm{Var}(T)=&\,12(11)+25(3)=207. \end{align*}

Hence T∼N(986,207)T\sim\mathrm{N}(986,207). Hence

P(T≤1000)= Φ(1000−986207)= 0.835(3 s.f.).\begin{align*} \mathrm{P}(T\leq1000)=&\,\Phi\bigg(\frac{1000-986}{\sqrt{207}}\bigg)\\[4mm] =&\,\boxed{0.835}\quad\text{(3 s.f.).} \end{align*}